Proof verification: ideal of $\mathbb{C}[x, y]$ generated by x, y is not a free module Let $R$ be $\mathbb{C}[x, y]$ and let $M$ be the ideal of $R$ generated by the two elements $x$ and $y$. Prove or disprove: $M$ is a free R-module.
This problem comes from Artin First Edition. It is problem 12.2.1. I am working on some problems near it 12.2.(4-7) as part of an informal study group, but this problem itself is excluded from the problem set.
This problem is similar to this one, but is not a strict duplicate because the answer to that question assumes a result that is proven later on in the problems. I wrote the solution to this problem before reading that answer.
Is my attempted solution to this problem correct? Is there a way to make the same argument that's simpler without appealing to more general results about $K[x, y]$?

I think this the statement is false and here is my attempt to prove it.
First a word on notation:
I'm going to use the following highly nonstandard notation.
Let $\mathop{\mu} \begin{bmatrix} S \cr R \end{bmatrix} $ denote the free R-module on S. Equivalently, $\mathop{\mu} \begin{bmatrix} S \cr R \end{bmatrix} $ is the set of formal linear combinations of $S$ with coefficients taken from $R$ and the symbols $\{0, +, -, *, =\}$ defined in appropriate way for formal linear combinations. It is a priori a free module.
Let $\mathop{\mu^e} \begin{bmatrix} S \cr R \end{bmatrix} $ denote the module derived from the free R-module on S by evaluating the formal linear combinations and identifying elements that are equal.
In prose, $X$ contains $Y$ means $X \in Y$. $X$ contains $Y$ by inclusion means $X \subset Y$.

Lemma: No element of $\mathop{\mu^e} \begin{bmatrix} \{x, y\} \cr \mathbb{C}[x, y] \end{bmatrix}$ has a non-zero constant term.
Proof: For any formal linear combination of elements of $\{x, y\}$ with coefficients in $\mathbb{C}[x, y]$, each of the terms has a zero constant subterm. Therefore, there is no way of producing an element of $\mathbb{C}[x, y]$ with a non-zero constant term in $\mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix}$.

We are asking ourselves, is $\mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix}$ free?
Equivalently, does there exist some subset $S$ of $\mathbb{C}[x,y]$ such that
$$ \mathop{\mu ^ e}\begin{bmatrix} \{x, y\} \cr \mathbb{C}[x,y] \end{bmatrix} \;\;\text{is isomorphic to}\;\; \mathop{\mu}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $$
My claim is no.
Suppose $0$ is an element of $S$, then consider $S'$ with $0$ removed.
Suppose $S$ has exactly two elements, $\alpha$ and $\beta$.
$ \mathop{\mu^e}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $ is not free because $0$ has a nontrivial representation before the evaluation step is applied.
$$ (-\beta) \cdot \alpha + (\alpha) \cdot \beta = 0 $$
Since $0$ has a nontrivial representation, that means that $ \mathop{\mu}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix} $ has stray equalities when evaluation is considered.
Suppose $S$ has more than two elements. Pick any two elements and apply the argument for the 2-element case.
Suppose $S$ has one element. Let $\cup S$ denote the sole element of $S$, borrowing notation from set theory.
$\cup S$ cannot be zero. $\cup S$ also cannot be a non-zero constant. If $\cup S$ were a non-zero constant, then we would be able to generate $2$ as an element of $\mathop{\mu^e}\begin{bmatrix} S \cr \mathbb{C}[x,y] \end{bmatrix}$, which contradicts the lemma.
$\cup S$ cannot have degree 2 or greater because then it would be unable to generate $x$.
$\cup S$ cannot have degree 1 because it must be able to generate both $x$ and $y$ and they have no common factors of degree 1.
$S$ cannot be the empty set because then the resulting module would not contain $x$.
 A: Since you show yourself quite open to simpler alternative approaches, may I try and present you with one, the inspiration for which comes from one of the exercises of the "Algèbre" of Bourbaki.
Consider a nonzero commutative ring $A$, the two indeterminate polynomial ring $B=A[X, Y]$ and the ideal $I\colon=BX+BY$. We aim to show that the $B$-module $I$ is not free, and we first claim that in order to prove this it will suffice to establish that in the tensor product (square) $E\colon=I \otimes_B I$ the element $X \otimes Y-Y \otimes X \neq 0_E$ is not null, however $XY(X \otimes Y-Y \otimes X)=(YX)\otimes(XY)-(XY)\otimes(YX)=0_E$.
Indeed, having established the above claim regarding the mentioned element of the tensor product let us assume by contradiction that $I$ were free, in other words $I \approx B^{(T)} \hspace{3pt} (\operatorname{\mathit{B}-\mathbf{Mod}})$ for a certain set $T$. Since $B$ is commutative, it would follow that the tensor product $E=I \otimes_B I \approx B^{(T)} \otimes_B B^{(T)} \approx B^{(T \times T)} \hspace{3pt} (\operatorname{\mathit{B}-\mathbf{Mod}})$ is also free. In general, for an arbitrary ring $C$ and a free (more generally, projective) left $C$-module $L$, it can be easily shown that given element $x \in L$ and operator $\lambda \in C$ which is not a left zero-divisor (this means that $\lambda$ is left simplifiable, or that $\lambda \mu=0_C \Rightarrow \mu=0_C$ for any $\mu \in C$), the condition $\lambda \in \mathrm{Ann}_C(x)$ entails $x=0_L$. In our particular instance, we have that $XY \in \mathrm{Ann}_B(X \otimes Y-Y \otimes X)$ and $XY$ is indeed a simplifiable element of $B$, however $X \otimes Y-Y \otimes X \neq 0_E$, which constitutes a contradiction.
Now on to justifying the claim concerning the tensor product. Consider the unique surjective unitary $A$-algebra morphism $\sigma \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Alg_1}}}(B, A)$ given by $\sigma(X)=\sigma(Y)=0_A$ -- this is nothing else than the morphism of substitution, taking any polynomial $f \in B$ to $f(0_A, 0_A)$ -- and consider ${}_{[\sigma]}A$, the $B$-module structure on $A$ obtained by scalar restriction via $\sigma$ from the canonical $A$-module structure on $A$. Explicitly, we are considering the $B$-module structure on $A$ for which $f.\lambda=f(0_A, 0_A)\lambda$ for any $f \in B$ and $\lambda \in A$.
Further consider the $B$-module $B^2$, with the canonical basis constituted by $e\colon=(1_A, 0_A)$ and $e'\colon=(0_A, 1_A)$ (we keep in mind that $0_B=0_A$ and $1_B=1_A$, since by construction $A$ is a subring of any ring of polynomials over it). We go on to define the unique $B$-bilinear map $\varphi \colon B^2 \times B^2 \to A$ such that $\varphi(e,e)=\varphi(e,e')=\varphi(e',e')=0_A$ and $\varphi(e',e)=1_A$. The general behaviour of this map is described by $\varphi\left((f,g), \left(f', g'\right)\right)=\left(gf'\right).1_A=\left(gf'\right)(0_A, 0_A)$.
Consider the $B$-module surjection $\rho \in \mathrm{Hom}_{\operatorname{\mathit{B}-\mathbf{Mod}}}\left(B^2, I\right)$ given by $\rho(e)=X$ and $\rho(e')=Y$. As Daniel Schepler pointed out in his above comment, it is easy to show that $\mathrm{Ker}\rho=B(Y, -X)$. We wish to remark here that the above bilinear map $\varphi$ is compatible with the surjection $\rho \times \rho \colon B^2 \times B^2 \to I \times I$, in the precise sense that $\rho(u)=\rho(u')$ and $\rho(v)=\rho(v')$ entail $\varphi(u, v)=\varphi(u',v')$ for any elements $u, u', v, v' \in B^2$. This is easily ascertained by remarking that $\varphi((Y, -X), u)=\varphi(u, (Y, -X))=0_A$ for any $u \in B^2$, relation which in turn stems from the fact that $X, Y \in \mathrm{Ann}_B\left({}_{[\sigma]}A\right)$. By virtue of a fundamental proposition in elementary set theory, the map $\varphi$ induces a unique quotient map $\psi \colon I \times I \to A$ such that $\varphi=\psi \circ (\rho \times \rho)$. This relation allows us to immediately ascertain the fact that $\psi$ is also $B$-bilinear and that $\psi(X, Y)=0_A$ respectively $\psi(Y, X)=1_A$.
By virtue of the universal property of tensor products, $\psi$ induces a unique $B$-linear map $\psi' \in \mathrm{Hom}_{\operatorname{\mathit{B}-\mathbf{Mod}}}(E, A)$ such that $\psi'(f \otimes g)=\psi(f, g)$ for any $f, g \in I$. It follows that $\psi'(X \otimes Y-Y \otimes X)=-1_A \neq 0_A$, whence necessarily $X \otimes Y-Y \otimes X \neq 0_E$.
The same argument can be used to show that $I$ is neither projective nor flat as a $B$-module (both these statements being stronger than the claim that $I$ is not free).

Before concluding, I would like to remark that the answer you included a link of in your posting is by far the most simple and efficient in the particular case when $A$ is a domain. The fact that $(X, Y)$ is not principal in $A[X, Y]$ can be immediately established.
